Find the larger of two positive real numbers that differ by 3 and have a product of 9. (If necessary, please r

Answer 1

Let x be your number. We know that the product of it and a number three smaller, is 9:

x(x-3) = 9
x^2 - 3x - 9 = 0

Solved using the quadratic formula:

x = (3 +/- sqrt(9 + 36))/2
x = (3 +/- 3 sqrt(5)) / 2

x = (3 + 3sqrt(5))/2, (3 - 3sqrt(5))/2

The answer they're looking for is the first one (3 + 3sqrt(5))/2, because the requirement is "two positive real numbers" and the other root is negative.

Checking the answer out:

x * (x -3)
(3 + 3sqrt(5))/2 * [(3 + 3sqrt(5))/2 - 3]
(3sqrt(5) + 3)/2 * [(3sqrt(5) - 3)/2]
([3sqrt(5)]^2 + 3*3sqrt(5) - 3*3sqrt(5) - 3^2)/4
(9*5 - 9)/4
36/4
9

Answer 2

You have to apply the quadratic equation solution. If it is will give you the real positive numbers. How come nobody is saying what a and b or x and y are ?

Check it out yourself.

Quadratic equation solutions:

http://www.jamesbrennan.org/algebra/quadratics/the_quadratic_formula.htm

Answer 3

a - b = 3
ab = 9

a = 3 + b
(3 + b)b = 9
3b + b^2 = 9
b^2 + 3b - 9 = 0
b = (-3 +/- sqrt(45)) / 2
b = (-3 +/- 3 sqrt 5) / 2
The answer is (-3 + 3 sqrt 5) / 2, if I'm right.